Proof of Cell Integrals of the Trigonometric Monomials

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. The field axioms of $\mathbb{R}$ are used freely for associativity, commutativity and distributivity, for $s\cdot1=s$, and for $ss^{-1}=1$ when $s\ne0$. Transitivity of $\le$ is used freely: if $a\le b$ and $b\le c$ then $0\le b-a$ and $0\le c-b$ by claim 3 of Elementary Arithmetic in an Ordered Field, hence $0\le c-a$ by claim 2 of that lemma, hence $a\le c$.

Notation. For $m\in\mathbb{Z}$ put $\omega_{m}=2\pi m$ and let $\gamma_{m}:\mathbb{R}\to\mathbb{R}$ be the map $\gamma_{m}(t)=\omega_{m}t$, so that $C_{m}=\cos\circ\gamma_{m}$ and $S_{m}=\sin\circ\gamma_{m}$.

Claim 1. (Elementary facts about the frequencies.) $0<\pi$; for $m\in\mathbb{Z}$ one has $\omega_{m}=0$ if and only if $m=0$; and for $k,m\in\mathbb{Z}$ the real numbers $k+m$, $k-m$ and $-m$ lie in $\mathbb{Z}$ with $\omega_{k}+\omega_{m}=\omega_{k+m}$ and $\omega_{k}-\omega_{m}=\omega_{k-m}$.

Proof of Claim 1. The real number $x_{0}$ of The Least Positive Zero of the Cosine §least-zero satisfies $0<x_{0}$, and $0<2$ by claim 8 of Elementary Order Arithmetic in an Ordered Field; hence $0<2x_{0}=\pi$ by claim 5 of Elementary Order Arithmetic in an Ordered Field and The Number Pi §pi. In particular $2\ne0$ and $\pi\ne0$. If $m=0$ then $\omega_{m}=2\pi\cdot0=0$ by claim 1 of Zero Products and Elementary Identities in a Field; if $m\ne0$ then $\omega_{m}\ne0$ by claim 3 of Zero Products and Elementary Identities in a Field, applied twice. Membership of $k+m$, $k-m$ and $-m$ in $\mathbb{Z}$ is claim 2 of Arithmetic, Order and Discreteness of the Integers. Finally $\omega_{k}+\omega_{m}=2\pi k+2\pi m=2\pi(k+m)=\omega_{k+m}$ by distributivity, and $\omega_{k}-\omega_{m}=2\pi k+(-(2\pi m))=2\pi k+2\pi(-m)=2\pi(k-m)=\omega_{k-m}$, using claim 2 of Zero Products and Elementary Identities in a Field for $-(2\pi m)=2\pi(-m)$ and the convention $x-y=x+(-y)$ of The Real Numbers: Standing Notation and Background §numbers. This proves Claim 1.

Proof of claim 1 of the statement. Fix $m\in\mathbb{Z}$. By claim 1 of Derivative of a Polynomial Function on the Real Line, applied with the interval $\mathbb{R}$, the restriction to $\mathbb{R}$ of the map $t\mapsto t^{1}$ is differentiable at every point with derivative $1$; and $t^{1}=t$ by claim 1 of Properties of Natural Number Powers in a Field, so this map is the identity map of $\mathbb{R}$. By claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, applied to the identity map with the constant $\omega_{m}$, the map $\gamma_{m}$ is differentiable at every $t\in\mathbb{R}$ with $\gamma_{m}'(t)=\omega_{m}\cdot1=\omega_{m}$.

By Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine §derivative the map $\cos$ is differentiable at every real number $x$ with derivative $-\sin x$, and $\sin$ is differentiable at $x$ with derivative $\cos x$. Apply Chain Rule for One-Dimensional Derivatives with both intervals equal to $\mathbb{R}$, with $\gamma=\gamma_{m}$ and with $g=\cos$, at the point $t$; every point of $\mathbb{R}$ is an interior point of $\mathbb{R}$ by Basic Facts about Intervals of the Real Line and Their Interior Points §whole-line. It gives that $C_{m}=\cos\circ\gamma_{m}$ is differentiable at $t$ with

using claim 2 of Zero Products and Elementary Identities in a Field for $(-a)b=-(ab)$ and commutativity. The same application with $g=\sin$ gives that $S_{m}$ is differentiable at $t$ with $S_{m}'(t)=\cos(\gamma_{m}(t))\,\omega_{m}=\omega_{m}C_{m}(t)$. These are the two asserted formulas.

Since $C_{m}$ and $S_{m}$ are differentiable at every point of $\mathbb{R}$, and every point of $\mathbb{R}$ is an interior point of $\mathbb{R}$, Differentiability at an Interior Point Implies Continuity There gives that both are continuous at every point of $\mathbb{R}$, hence continuous on $\mathbb{R}$. The bounds $|C_{m}(t)|\le1$ and $|S_{m}(t)|\le1$ are The Pythagorean Identity for Sine and Cosine §bounds, applied at the real number $\gamma_{m}(t)$.

By Claim 1, $\omega_{0}=0$, so $\gamma_{0}(t)=0\cdot t=0$ for every $t$ by claim 1 of Zero Products and Elementary Identities in a Field; hence $C_{0}(t)=\cos0=1$ and $S_{0}(t)=\sin0=0$ by Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine §values.

For $k,m\in\mathbb{Z}$ the products $C_{k}C_{m}$, $S_{k}S_{m}$ and $S_{k}C_{m}$ are continuous on $\mathbb{R}$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, applied on the set $\mathbb{R}$ to the two continuous companion maps in each case. Each of the maps $C_{m}$, $S_{m}$, $C_{k}C_{m}$, $S_{k}S_{m}$, $S_{k}C_{m}$ is therefore a continuous map from $\mathbb{R}$ to $\mathbb{R}$, so Zero Extension of a Real-Valued Function, and the Unit-Cell Integral of a Continuous Function §continuous gives that its restriction to $J$ is $\mathcal{B}_{J}$-measurable and $\lambda_{J}$-integrable, and that its integral over $J$ equals the Riemann integral over $[0,1]$ of its restriction to $[0,1]$.

Proof of claim 2 of the statement. Suppose first that $m=0$. The restriction $C_{0}|_{J}$ is the map with constant value $1$ on $J$, which is the indicator $\mathbf{1}_{J}$ of $J$ inside the measure space $(J,\mathcal{B}_{J},\lambda_{J})$; so $\int_{J}C_{0}\,d\lambda_{J}=\lambda_{J}(J)=1$ by The Integral of an Indicator Function is the Measure of the Set and The Integral over the Unit Cell of a Product of One-Variable Functions, where $\lambda(J)=1$ is recorded. The restriction $S_{0}|_{J}$ is the map with constant value $0$, which is $0\cdot\mathbf{1}_{J}$; by the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral, taken with the constant $0$ and the companion $\mathbf{1}_{J}$, its integral is $0\cdot1=0$.

Suppose now that $m\ne0$, so that $\omega_{m}\ne0$ by Claim 1 and $\omega_{m}^{-1}$ exists. Let $A_{m}=\omega_{m}^{-1}S_{m}$ and $B_{m}=-\omega_{m}^{-1}C_{m}$, that is, the maps $t\mapsto\omega_{m}^{-1}S_{m}(t)$ and $t\mapsto(-\omega_{m}^{-1})C_{m}(t)$. By claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, applied to the companion $S_{m}$ with the constant $\omega_{m}^{-1}$ and to the companion $C_{m}$ with the constant $-\omega_{m}^{-1}$, together with claim 1 of the statement, these maps are differentiable at every $t\in\mathbb{R}$ with

the second using claim 2 of Zero Products and Elementary Identities in a Field for $(-x)(-y)=xy$. In particular $A_{m}$ and $B_{m}$ are continuous on $\mathbb{R}$ by Differentiability at an Interior Point Implies Continuity There.

Let $[0,1]$ be the closed interval determined by $0<1$ and write $(0,1)=\{x\in\mathbb{R}:0<x<1\}$. By claim 1 of Restriction Stability of Continuity and of the Derivative the restrictions $C_{m}|_{[0,1]}$, $S_{m}|_{[0,1]}$, $A_{m}|_{[0,1]}$ and $B_{m}|_{[0,1]}$ are continuous on $[0,1]$, and by A Continuous Function on a Closed Interval is Riemann Integrable §integrable the first two are Riemann integrable on $[0,1]$. Every $x\in(0,1)$ is an interior point of $[0,1]$, as recorded in Fundamental Theorem of Calculus, Part II, on a Closed Real Interval, so by claim 2 of Restriction Stability of Continuity and of the Derivative, applied with the intervals $[0,1]\subseteq\mathbb{R}$, the restriction $A_{m}|_{[0,1]}$ is differentiable at $x$ with derivative $A_{m}'(x)=C_{m}(x)$, and likewise $(B_{m}|_{[0,1]})'(x)=S_{m}(x)$. Hence Fundamental Theorem of Calculus, Part II, on a Closed Real Interval, applied on $[0,1]$ with $f=C_{m}|_{[0,1]}$ and $F=A_{m}|_{[0,1]}$ and then with $f=S_{m}|_{[0,1]}$ and $F=B_{m}|_{[0,1]}$, gives

Now $\gamma_{m}(1)=\omega_{m}=2\pi m$ and $\gamma_{m}(0)=0$, so by Quarter-Turn Identities and Periodicity of Sine and Cosine §integer one has $S_{m}(1)=\sin(2\pi m)=0$ and $C_{m}(1)=\cos(2\pi m)=1$, while $S_{m}(0)=\sin0=0$ and $C_{m}(0)=\cos0=1$ by Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine §values. Therefore $A_{m}(1)-A_{m}(0)=\omega_{m}^{-1}(0-0)=0$ and $B_{m}(1)-B_{m}(0)=(-\omega_{m}^{-1})(1-1)=0$, both by claim 1 of Zero Products and Elementary Identities in a Field. By the last sentence of the proof of claim 1 of the statement, $\int_{J}C_{m}\,d\lambda_{J}$ and $\int_{J}S_{m}\,d\lambda_{J}$ equal these two Riemann integrals, hence both vanish. Together with the case $m=0$ this proves claim 2 of the statement.

Claim 2. (Equality of absolute values.) For $k,m\in\mathbb{Z}$ one has $|k|=|m|$ if and only if $k=m$ or $k=-m$; and if $k\ne0$ then not both.

Proof of Claim 2. Suppose $|k|=|m|$. By claim 1 of Nonnegativity of Squares in an Ordered Field, $kk=|k|\,|k|=|m|\,|m|=mm$, so $(k-m)(k+m)=kk-mm=0$ by claim 4 of Zero Products and Elementary Identities in a Field; hence $k-m=0$ or $k+m=0$ by claim 3 of that lemma, that is, $k=m$ or $k=-m$. Conversely, if $k=m$ then $|k|=|m|$, and if $k=-m$ then $|k|=|-m|=|m|$ by claim 2 of Properties of the Absolute Value in an Ordered Field. Finally, if $k=m$ and $k=-m$ then $k+k=m+(-m)=0$, so $2k=0$ and hence $k=0$ by claim 3 of Zero Products and Elementary Identities in a Field, since $2\ne0$ by Claim 1. This proves Claim 2.

Claim 3. (The three product identities, integrated.) For all $k,m\in\mathbb{Z}$,

Proof of Claim 3. Fix $k,m\in\mathbb{Z}$ and $t\in\mathbb{R}$, and put $u=\gamma_{k}(t)$ and $v=\gamma_{m}(t)$. By Claim 1 and distributivity, $u+v=(\omega_{k}+\omega_{m})t=\omega_{k+m}t=\gamma_{k+m}(t)$ and $u-v=(\omega_{k}-\omega_{m})t=\gamma_{k-m}(t)$, and $k+m,k-m\in\mathbb{Z}$. Hence Product-to-Sum Formulas for Sine and Cosine §cosine-cosine, Product-to-Sum Formulas for Sine and Cosine §sine-sine and Product-to-Sum Formulas for Sine and Cosine §sine-cosine, applied at these $u$ and $v$, read

Each of the seven maps appearing here has a $\lambda_{J}$-integrable restriction to $J$, by claim 1 of the statement. Restricting the first identity to $J$ and applying claim 2 of Linearity and Monotonicity of the Lebesgue Integral, first with the constants $1$ and $1$ and the companions $C_{k-m}|_{J}$ and $C_{k+m}|_{J}$, and then with the constants $2$ and $0$ and the companion $(C_{k}C_{m})|_{J}$ in both slots, gives the first displayed identity of the claim. The second identity follows in the same way, taking the constants $1$ and $-1$ in the first application, and the third likewise from the third pointwise identity. This proves Claim 3.

Proof of claim 3 of the statement. Let $k,m\in\mathbb{Z}$ and write $P=\int_{J}C_{k-m}\,d\lambda_{J}+\int_{J}C_{k+m}\,d\lambda_{J}$, so that $\int_{J}C_{k}C_{m}\,d\lambda_{J}=2^{-1}P$ by Claim 3.

If $k=0$ and $m=0$ then $k-m=0$ and $k+m=0$, so $P=1+1=2$ by claim 2 of the statement and $2^{-1}P=1$.

Suppose $k\ne0$ and $|k|=|m|$. By Claim 2 exactly one of $k=m$ and $k=-m$ holds. If $k=m$ then $k-m=0$ and $k+m=k+k=2k$, which is nonzero because $2\ne0$ and $k\ne0$ (claim 3 of Zero Products and Elementary Identities in a Field); so $P=1+0=1$ by claim 2 of the statement. If $k=-m$ then $k+m=0$ and $k-m=k+k=2k\ne0$, so $P=0+1=1$. In both cases $2^{-1}P=2^{-1}$.

Suppose finally $|k|\ne|m|$. By Claim 2, $k\ne m$ and $k\ne-m$, so $k-m\ne0$ and $k+m\ne0$, whence $P=0+0=0$ and $2^{-1}P=0$. These three cases are exhaustive, since $|k|=|m|$ together with $k=0$ forces $|m|=|0|=0$, and hence $m=0$, by claim 1 of Properties of the Absolute Value in an Ordered Field.

Proof of claim 4 of the statement. Let $k,m\in\mathbb{Z}$ and write $D=\int_{J}C_{k-m}\,d\lambda_{J}-\int_{J}C_{k+m}\,d\lambda_{J}$, so that $\int_{J}S_{k}S_{m}\,d\lambda_{J}=2^{-1}D$ by Claim 3.

If $k\ne0$ and $k=m$ then, as above, $k-m=0$ and $k+m=2k\ne0$, so $D=1-0=1$ and $2^{-1}D=2^{-1}$. If $k\ne0$ and $k=-m$ then $k+m=0$ and $k-m=2k\ne0$, so $D=0-1=-1$ and $2^{-1}D=-2^{-1}$.

If $k=0$ then $k-m=-m$ and $k+m=m$. When $m=0$ both are $0$ and $D=1-1=0$; when $m\ne0$ both are nonzero, by Claim 1 applied to $-m$ and to $m$ together with claim 2 of Zero Products and Elementary Identities in a Field, and $D=0-0=0$. If $|k|\ne|m|$ then, by Claim 2, $k-m\ne0$ and $k+m\ne0$, so $D=0-0=0$. In both of these remaining cases $2^{-1}D=0$.

Proof of claim 5 of the statement. Let $k,m\in\mathbb{Z}$. By claim 2 of the statement, $\int_{J}S_{k+m}\,d\lambda_{J}=0$ and $\int_{J}S_{k-m}\,d\lambda_{J}=0$, so the third identity of Claim 3 gives $2\int_{J}S_{k}C_{m}\,d\lambda_{J}=0$ and hence $\int_{J}S_{k}C_{m}\,d\lambda_{J}=0$, since $2\ne0$ and claim 3 of Zero Products and Elementary Identities in a Field applies.